Paul's Online Notes
Paul's Online Notes
Home / Algebra / Graphing and Functions / Circles
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 3-3 : Circles

6. Determine the radius and center of the following circle. If the equation is not the equation of a circle clearly explain why not.

\[{x^2} + {y^2} + 14x - 8y + 56 = 0\]

Show All Steps Hide All Steps

Start Solution

To do this problem we need to complete the square on the \(x\) and \(y\) terms. To help with this we’ll first get the number on the right side and group the \(x\) and \(y\) terms as follows.

\[{x^2} + 14x + {y^2} - 8y = - 56\] Show Step 2

Here are the numbers we need to complete the square for both \(x\) and \(y\).

\[x:\,{\left( {\frac{{14}}{2}} \right)^2} = {\left( 7 \right)^2} = 49\hspace{0.5in}y:\, {\left( {\frac{{ - 8}}{2}} \right)^2} = {\left( { - 4} \right)^2} = 16\] Show Step 3

Now, complete the square.

\[\begin{align*}{x^2} + 14x + 49 + {y^2} - 8y + 16 & = - 56 + 49 + 16\\ {\left( {x + 7} \right)^2} + {\left( {y - 4} \right)^2} & = 9\end{align*}\]

Don’t forget to add the numbers from Step 2 to both sides of the equation!

Show Step 4

So, we have the equation in standard form and so we can quickly identify the radius and center of the circle.

\[{\mbox{Radius : }}r = 3\hspace{0.25in}\hspace{0.25in}{\mbox{Center : }}\left( { - 7,4} \right)\]

If you don’t recall how to get the radius and center from the standard form of a circle check out Problems 3 – 5 in this section for some practice.